posted May 14, 2003 02:21 PM               

Of course, the wider the flight, the less space there is for material, so therefore one would expect less amperage. Think of a screw that was all flight and no material . . . no amps there!

So my answer is the wider load bearing surface would share the wear and thus "wear less"

posted May 14, 2003 03:27 PM               

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You can also make the flight pitch greater

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or the screw deeper to compensate for the lost volume

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I'm not saying you can't have a wider flight, just that there may be implications to think about if you do. Another potential problem could be the amount of shear the material that passes over the top of a wider flight would be exposed to.A wider flight will increase the amount of shear and could cause a gel problem, a rule of thumb for flight width is 10% of the diameter.

I agree with Louis that motor amps will go down, this would be expected if less material is being moved by the screw.

posted May 14, 2003 03:47 PM               

Other thoughts;

A wider flight will make a stiffer screw, less subject to local deflection, and therefore speading the load over an even greater bearing surface.

I beleive the optimal solution to all of these issues had been found in the double flighted screw.

posted May 15, 2003 07:25 AM               

There is an argument that says that approximately 40% of the motor load is due to material leaking over the flight land and that a narrow flight width is advantageous. However, as a practical matter, a flight width which is much less than 10% of the screw diameter lacks structural integrity and results in flight fracture. Experience has shown that the 10% rule works well...

posted May 15, 2003 05:56 PM               

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There is an argument that says that approximately 40% of the motor load is due to material leaking over the flight land and that a narrow flight width is advantageous. However, as a practical matter, a flight width which is much less than 10% of the screw diameter lacks structural integrity and results in flight fracture. Experience has shown that the 10% rule works well...

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So then a barrier type screw would have a greater load at the same pounds per hour?

posted May 16, 2003 08:05 AM               

That's an interesting question. As the motor power is the shear force X rotational velocity of the solid bed, melt pool and screw flight respectively, the additional barrier flight would be included in this expression. Moreover, the shear force is the shear stress X area of each of the contributing components. As such, given a barrier flight which may be as little as 5% of the screw lead, there may still be a significant contribution to motor load based on the shear stress through this narrow gap. You would have to evaluate the viscosity decay due to shear rate based on the power law value of the fluid. Depending on the decay of the fluid, the shear stress may be very high in this component resulting in a significant contribution to motor load.